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Integrate the function\[\frac{x^3\sin\bigg(\tan^{-1}x^4\bigg)}{1+x^8}\]

$\begin{array}{1 1}\large \frac{-\cos(\tan^{-1}x^4)}{4}+c \\ \large \frac{\cos(\tan^{-1}x^4)}{4}+c \\ \large \frac{-\sin (\tan^{-1}x^4)}{4}+c \\ \large \frac{\sin (\tan^{-1}x^4)}{4}+c \end{array}$

1 Answer

  • (i)Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • $\Rightarrow $dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
  • (ii)$\int \frac{1}{x}dx=log x+c.$
Given $I=\int \frac{x^3\sin (\tan^{-1}x^4)}{1+x^8}dx$.
Put $\tan^{-1}x^4=t$
$\Rightarrow \frac{x^3}{1+x^8}dx=\frac{dt}{4}.$
Now substituting of x and dx we get,
$I=\int \frac{dt}{4}.t\sin t=\frac{1}{4}\int\sin t.dt.$
On integrating we get,
Substituting back for t we get,
$\int \frac{x^3\sin (\tan^{-1}x^4)}{1+x^8}dx=\frac{-1}{4}\cos(\tan^{-1}x^4)+c.$


answered Jan 30, 2013 by sreemathi.v